Simplify the following expression: $y = \dfrac{2x^2- 5x- 25}{x - 5}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(-25)} &=& -50 \\ {a} + {b} &=& &=& {-5} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-50$ and add them together. Remember, since $-50$ is negative, one of the factors must be negative. The factors that add up to ${-5}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${-10}$ $ \begin{eqnarray} {ab} &=& ({5})({-10}) &=& -50 \\ {a} + {b} &=& {5} + {-10} &=& -5 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 +{5}x) + ({-10}x {-25}) $ Factor out the common factors: $ x(2x + 5) - 5(2x + 5)$ Now factor out $(2x + 5)$ $ (2x + 5)(x - 5)$ The original expression can therefore be written: $ \dfrac{(2x + 5)(x - 5)}{x - 5}$ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ This leaves us with $2x + 5; x \neq 5$.